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Q.

A ball weighing 10 g hits a hard surface vertically with a speed of 5 m/s and rebounds with the same speed. The ball remains in contact with the surface for 0.01 sec. The average force exerted by the surface on the ball is:

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a

1 N

b

10 N

c

100 N

d

0.1N

answer is B.

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Detailed Solution

$F_{av} = \frac{Δp}{Δt} = \frac{(10 \times 10^{- 3}) (5 + 5)}{0 .01} = 10 N$

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